Solve x/x+4=10/x

$$\frac { x } { x + 4 } = \frac { 10 } { x }$$

Solve for x

$$x=\sqrt{65}+5\approx 13.062257748$$

$$x=5-\sqrt{65}\approx -3.062257748$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x+4,x.

$$xx=\left(x+4\right)\times 10$$

Multiply x and x to get x^{2}.

$$x^{2}=\left(x+4\right)\times 10$$

Use the distributive property to multiply x+4 by 10.

$$x^{2}=10x+40$$

Subtract 10x from both sides.

$$x^{2}-10x=40$$

Subtract 40 from both sides.

$$x^{2}-10x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-40\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$x=\frac{-\left(-10\right)±\sqrt{100+160}}{2}$$

Add 100 to 160.

$$x=\frac{-\left(-10\right)±\sqrt{260}}{2}$$

Take the square root of 260.

$$x=\frac{-\left(-10\right)±2\sqrt{65}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{65}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{65}}{2} when ± is plus. Add 10 to 2\sqrt{65}.

$$x=\frac{2\sqrt{65}+10}{2}$$

Divide 10+2\sqrt{65} by 2.

$$x=\sqrt{65}+5$$

Now solve the equation x=\frac{10±2\sqrt{65}}{2} when ± is minus. Subtract 2\sqrt{65} from 10.

$$x=\frac{10-2\sqrt{65}}{2}$$

Divide 10-2\sqrt{65} by 2.

$$x=5-\sqrt{65}$$

The equation is now solved.

$$x=\sqrt{65}+5 x=5-\sqrt{65}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x+4,x.

$$xx=\left(x+4\right)\times 10$$

Multiply x and x to get x^{2}.

$$x^{2}=\left(x+4\right)\times 10$$

Use the distributive property to multiply x+4 by 10.

$$x^{2}=10x+40$$

Subtract 10x from both sides.

$$x^{2}-10x=40$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=40+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=40+25$$

Add 40 to 25.

$$x^{2}-10x+25=65$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=65$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{65}$$

Simplify.

$$x-5=\sqrt{65} x-5=-\sqrt{65}$$

Add 5 to both sides of the equation.

$$x=\sqrt{65}+5 x=5-\sqrt{65}$$